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Statistics question

2021 · 16 Mar · Shift 1 · Q32
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  5. /2021 · 16 Mar · Shift 1 · Q32

Statistics question

2021 · 16 Mar · Shift 1 · Q32

JEE MainMathematicsStatisticsMCQ+4 / −1
Consider three observations a, b, and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?
  1. A
    b2 = 3(a2 + c2) + 9d2
  2. B
    b2 = 3(a2 + c2) −-− 9d2
  3. C
    b2 = 3(a2 + c2 + d2)
  4. D
    b2 = a2 + c2 + 3d2
View written solutionFree

Correct answer: B

  1. Use the property of standard deviation under translation

If the standard deviation of a+2, b+2, c+2a+2,\ b+2,\ c+2a+2, b+2, c+2 is ddd, then adding the same constant to all observations does not change the standard deviation.

So the standard deviation of a,b,ca, b, ca,b,c is also ddd.

  1. Write the mean of the three observations

Given: b=a+cb=a+cb=a+c

Mean of a,b,ca,b,ca,b,c is xˉ=a+b+c3\bar{x}=\frac{a+b+c}{3}xˉ=3a+b+c​

Substitute b=a+cb=a+cb=a+c: xˉ=a+(a+c)+c3=2a+2c3=2(a+c)3=2b3\bar{x}=\frac{a+(a+c)+c}{3}=\frac{2a+2c}{3}=\frac{2(a+c)}{3}=\frac{2b}{3}xˉ=3a+(a+c)+c​=32a+2c​=32(a+c)​=32b​

  1. Use the formula for standard deviation

For three observations, variance is d2=13[(a−xˉ)2+(b−xˉ)2+(c−xˉ)2]d^2=\frac{1}{3}\left[(a-\bar{x})^2+(b-\bar{x})^2+(c-\bar{x})^2\right]d2=31​[(a−xˉ)2+(b−xˉ)2+(c−xˉ)2]

Since xˉ=2b3\bar{x}=\frac{2b}{3}xˉ=32b​, d2=13[(a−2b3)2+(b−2b3)2+(c−2b3)2]d^2=\frac{1}{3}\left[\left(a-\frac{2b}{3}\right)^2+\left(b-\frac{2b}{3}\right)^2+\left(c-\frac{2b}{3}\right)^2\right]d2=31​[(a−32b​)2+(b−32b​)2+(c−32b​)2]

Now use b=a+cb=a+cb=a+c.

Then a−2b3=a−2(a+c)3=a−2c3a-\frac{2b}{3}=a-\frac{2(a+c)}{3}=\frac{a-2c}{3}a−32b​=a−32(a+c)​=3a−2c​ b−2b3=b3b-\frac{2b}{3}=\frac{b}{3}b−32b​=3b​ c−2b3=c−2(a+c)3=c−2a3c-\frac{2b}{3}=c-\frac{2(a+c)}{3}=\frac{c-2a}{3}c−32b​=c−32(a+c)​=3c−2a​

So, d2=13[(a−2c3)2+(b3)2+(c−2a3)2]d^2=\frac{1}{3}\left[\left(\frac{a-2c}{3}\right)^2+\left(\frac{b}{3}\right)^2+\left(\frac{c-2a}{3}\right)^2\right]d2=31​[(3a−2c​)2+(3b​)2+(3c−2a​)2]

d2=127[(a−2c)2+b2+(c−2a)2]d^2=\frac{1}{27}\left[(a-2c)^2+b^2+(c-2a)^2\right]d2=271​[(a−2c)2+b2+(c−2a)2]

  1. Expand and simplify

Expand: (a−2c)2=a2−4ac+4c2(a-2c)^2=a^2-4ac+4c^2(a−2c)2=a2−4ac+4c2 (c−2a)2=c2−4ac+4a2(c-2a)^2=c^2-4ac+4a^2(c−2a)2=c2−4ac+4a2

Thus, (a−2c)2+(c−2a)2=5a2+5c2−8ac(a-2c)^2+(c-2a)^2=5a^2+5c^2-8ac(a−2c)2+(c−2a)2=5a2+5c2−8ac

Also, b2=(a+c)2=a2+2ac+c2b^2=(a+c)^2=a^2+2ac+c^2b2=(a+c)2=a2+2ac+c2

Therefore,

= (5a^2+5c^2-8ac)+(a^2+2ac+c^2) =6a^2+6c^2-6ac$$ Hence, $$d^2=\frac{1}{27}(6a^2+6c^2-6ac)=\frac{2}{9}(a^2+c^2-ac)$$ So, $$9d^2=2(a^2+c^2-ac)$$ --- 5. **Relate this with $b^2$** Since $$b^2=(a+c)^2=a^2+2ac+c^2$$ Now compute: $$3(a^2+c^2)-9d^2$$ Substitute $9d^2=2(a^2+c^2-ac)$: $$3(a^2+c^2)-2(a^2+c^2-ac)$$ $$=3a^2+3c^2-2a^2-2c^2+2ac$$ $$=a^2+c^2+2ac$$ $$=(a+c)^2=b^2$$ Therefore, $$b^2=3(a^2+c^2)-9d^2$$ --- 6. **Check the options** - **A:** $b^2=3(a^2+c^2)+9d^2$ ❌ - **B:** $b^2=3(a^2+c^2)-9d^2$ ✅ - **C:** $b^2=3(a^2+c^2+d^2)$ ❌ - **D:** $b^2=a^2+c^2+3d^2$ ❌ So the correct option is **B**.
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